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3 years ago

The cart is initially at rest. Force

Physics

1 answer:

Oliga [24]3 years ago

5 0

Answer:

$v'=\dfrac{1}{2}v$

Explanation:

Given that,

Initial speed of the cart, u = 0

Let F force is applied to the cart for time $\Delta t$ after which the car has speed v. The force on an object is given by :

F = ma

m is the mass of the cart

We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,

$F=\dfrac{mv'}{t}$

$v'=\dfrac{F t}{2m}$

$v'=\dfrac{1}{2}v$

So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).

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a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force

loris [4]

Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

applied force on the plane, Fa = 706 N

the net force on the plane, ∑F= 450 N

Let the force due to air resistance = Fr

The net force on the plane is given as;

Net force = applied force - force due to air resistance

∑F = Fa - Fr

Fr = Fa - ∑F

Fr = 706 - 450

Fr = 256 N.

Therefore, the force due to air resistance is 256 N.

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3 years ago

What is menat by the age of the universe? how old is the universe?

ikadub [295]

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3 years ago

Read 2 more answers

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

8 0

1 year ago

An 800 kg fishing boat going south at 12 m/s runs into a stopped pontoon boat (1400 kg). The boats stick together and move south

kykrilka [37]

Answer:

the velocity of the boats after the collision is 4.36 m/s.

Explanation:

Given;

mass of fish, m₁ = 800 kg

mass of boat, m₂ = 1400 kg

initial velocity of the fish, u₁ = 12 m/s

initial velocity of the boat, u₂ = 0

let the final velocity of the fish-boat after collision = v

Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

800 x 12 + 1400 x 0 = v(800 + 1400)

9600 = 2200v

v = 9600/2200

v = 4.36 m/s

Therefore, the velocity of the boats after the collision is 4.36 m/s.

7 0

3 years ago